Balancing chemical equations worksheet with various reactions involving gases, solids, and aqueous solutions.

We are given several unbalanced chemical equations and need to balance them. Let’s go one by one, using the standard method: count atoms of each element on both sides and adjust coefficients.

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2. CH₄ + O₂ → CO₂ + H₂O
Left: C=1, H=4, O=2
Right: C=1, O=2+1=3 (CO₂ has 2 O, H₂O has 1 O), H=2

To balance H: CH₄ has 4 H, so need 2 H₂O → gives 4 H and 2 O from water.
Now right side: C=1, H=4, O = 2 (from CO₂) + 2 (from 2 H₂O) = 4 O
So need 2 O₂ on left (since each O₂ gives 2 O → 2×2 = 4 O)

Check:
CH₄ + 2O₂ → CO₂ + 2H₂O
C: 1 = 1 ✔
H: 4 = 2×2 = 4 ✔
O: 2×2 = 4; right: 2 (CO₂) + 2×1 = 4 ✔

✔ Balanced: 1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O

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3. Xe + F₂ → XeF₆
Xenon hexafluoride: one Xe, six F.
F₂ is diatomic, so need 3 F₂ to get 6 F atoms.

Xe + 3F₂ → XeF₆
Xe: 1 = 1 ✔
F: 3×2 = 6 = 6 ✔

✔ Balanced: 1 Xe + 3 F₂ → 1 XeF₆

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4. H₂SO₄ + Pb(OH)₂ → Pb(SO₄)₂ + H₂O
Wait — check formula: Lead(II) sulfate is PbSO₄, not Pb(SO₄)₂.
Pb is usually +2, SO₄ is -2 → PbSO₄.
But the problem writes Pb(SO₄)₂ — that would imply Pb⁴⁺, which is rare. However, since the problem gives Pb(SO₄)₂, we’ll assume it's intentional (maybe lead(IV) sulfate — though uncommon). Let’s proceed as written.

Left:
H₂SO₄: H=2, S=1, O=4
Pb(OH)₂: Pb=1, O=2, H=2
Total left: Pb=1, S=1, H=4, O=6

Right:
Pb(SO₄)₂: Pb=1, S=2, O=8
H₂O: H=2, O=1 per molecule

We need 2 S on right, but only 1 S on left → need 2 H₂SO₄
Try: 2 H₂SO₄ + Pb(OH)₂ → Pb(SO₄)₂ + ? H₂O

Left now:
H: 2×2 = 4 from H₂SO₄ + 2 from Pb(OH)₂ = 6 H
S: 2
O: 2×4 = 8 from H₂SO₄ + 2 from Pb(OH)₂ = 10 O
Pb: 1

Right: Pb(SO₄)₂ → Pb=1, S=2, O=8
Remaining H and O must go to H₂O.

We have 6 H on left → need 3 H₂O (each has 2 H) → 3 H₂O gives 6 H and 3 O
Then total right O = 8 (from sulfate) + 3 = 11 — but left has only 10 O. Not balanced.

Alternative: maybe the product is actually PbSO₄, not Pb(SO₄)₂. That makes more sense chemically:
H₂SO₄ + Pb(OH)₂ → PbSO₄ + 2 H₂O
Check:
Left: H=2+2=4, S=1, O=4+2=6, Pb=1
Right: Pb=1, S=1, O=4+2×1=6, H=2×2=4 ✔
Yes! This is correct and standard.

Given that the problem wrote Pb(SO₄)₂, but that leads to inconsistency unless Pb is +4 (which is unusual for this reaction), and since this is likely a high-school level problem, it's probably a typo, and should be PbSO₄.

But the instruction says: solve the problem *as given*. So let’s double-check if Pb(SO₄)₂ can be balanced.

Assume Pb(SO₄)₂ is correct → Pb⁴⁺, SO₄²⁻ ×2 → net 0.

Then we need 2 sulfate ions → 2 H₂SO₄
Pb(OH)₂ provides Pb²⁺, but we need Pb⁴⁺ — not matching. So chemically unlikely.

Given context, I believe it's a typo and should be PbSO₄. Many textbooks use this exact reaction: acid + base → salt + water.

So we’ll go with:
H₂SO₄ + Pb(OH)₂ → PbSO₄ + 2 H₂O

But the problem explicitly writes Pb(SO₄)₂. Hmm.

Let me search my knowledge: Is Pb(SO₄)₂ a real compound? Lead(IV) sulfate exists but decomposes easily and is not typical in intro chem.

Since this is a balancing exercise (not testing real chemistry), perhaps they just want atom balance regardless of realism.

Try balancing with Pb(SO₄)₂:

Let coefficients be: a H₂SO₄ + b Pb(OH)₂ → c Pb(SO₄)₂ + d H₂O

Atom counts:
Pb: b = c
S: a = 2c
H: 2a + 2b = 2d
O: 4a + 2b = 8c + d

From Pb: b = c
From S: a = 2c → a = 2b
Plug into H: 2(2b) + 2b = 2d → 4b + 2b = 6b = 2d → d = 3b
Now O: 4a + 2b = 4(2b) + 2b = 8b + 2b = 10b
Right: 8c + d = 8b + 3b = 11b
10b ≠ 11b → no integer solution.

So impossible to balance if product is Pb(SO₄)₂.

Therefore, the intended product is PbSO₄.

I will assume typo and use PbSO₄.

✔ Balanced: 1 H₂SO₄ + 1 Pb(OH)₂ → 1 PbSO₄ + 2 H₂O

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5. PCl₅ + H₂O → H₃PO₄ + HCl
Left: P=1, Cl=5, H=2, O=1
Right: H₃PO₄: H=3, P=1, O=4; HCl: H=1, Cl=1 → total H=4, Cl=1, O=4

Need 5 Cl on right → 5 HCl
Then H from HCl = 5, plus 3 from H₃PO₄ = 8 H needed on left
Each H₂O gives 2 H → need 4 H₂O
Check O: 4 H₂O → 4 O; right: H₃PO₄ has 4 O → OK
P: 1 = 1
Cl: 5 = 5
H: 4×2 = 8; right: 3 (H₃PO₄) + 5 (HCl) = 8 ✔

So: PCl₅ + 4 H₂O → H₃PO₄ + 5 HCl

✔ Balanced.

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6. KNO₃ + H₂CO₃ → K₂CO₃ + HNO₃
Left: K=1, N=1, O=3+3=6, H=2, C=1
Right: K₂CO₃: K=2, C=1, O=3; HNO₃: H=1, N=1, O=3 → total K=2, C=1, H=1, N=1, O=6

K mismatch: left 1, right 2 → need 2 KNO₃
Then left: K=2, N=2, O=2×3=6 + 3 = 9, H=2, C=1
Right: need 2 HNO₃ to match N=2 and H=2
Then right: K₂CO₃ (K=2, C=1, O=3) + 2 HNO₃ (H=2, N=2, O=6) → total O=3+6=9, H=2, K=2, C=1, N=2 ✔

So: 2 KNO₃ + H₂CO₃ → K₂CO₃ + 2 HNO₃

✔ Balanced.

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7. BF₃ + Li₂SO₃ → B₂(SO₃)₃ + LiF
Left: B=1, F=3, Li=2, S=1, O=3
Right: B₂(SO₃)₃ → B=2, S=3, O=9; LiF → Li=1, F=1

We need 2 B on right → 2 BF₃ on left
Then left: B=2, F=6, Li=2, S=1, O=3
Right: B₂(SO₃)₃ needs S=3, O=9 → so need 3 Li₂SO₃ on left
Try: 2 BF₃ + 3 Li₂SO₃ → B₂(SO₃)₃ + ? LiF

Left now:
B=2, F=6, Li=6, S=3, O=9
Right: B₂(SO₃)₃ → B=2, S=3, O=9
Remaining: Li=6, F=6 → need 6 LiF

So: 2 BF₃ + 3 Li₂SO₃ → B₂(SO₃)₃ + 6 LiF

Check:
B: 2=2 ✔
F: 2×3=6 = 6 ✔
Li: 3×2=6 = 6 ✔
S: 3=3 ✔
O: 3×3=9 = 9 ✔

✔ Balanced.

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8. Be₂C + H₂O → Be(OH)₂ + CH₄(aq)
Beryllium carbide reacts with water to give beryllium hydroxide and methane.

Left: Be=2, C=1, H=2, O=1
Right: Be(OH)₂: Be=1, O=2, H=2; CH₄: C=1, H=4 → total Be=1, C=1, H=6, O=2

Be mismatch: need 2 Be(OH)₂
Then right: Be=2, O=4, H=4 (from 2×Be(OH)₂) + 4 (from CH₄) = 8 H, C=1

Left: Be₂C has Be=2, C=1
Need H₂O to provide H and O. Each H₂O gives 2H, 1O.

We need 8 H → 4 H₂O → gives 8 H and 4 O
Left: Be₂C + 4 H₂O → 2 Be(OH)₂ + CH₄
Check atoms:
Be: 2 = 2 ✔
C: 1 = 1 ✔
H: 4×2 = 8; right: 2×2 = 4 from Be(OH)₂ + 4 from CH₄ = 8 ✔
O: 4×1 = 4; right: 2×2 = 4 from Be(OH)₂ ✔

✔ Balanced: 1 Be₂C + 4 H₂O → 2 Be(OH)₂ + 1 CH₄

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Now compile all balanced equations with coefficients:

2. 1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O
3. 1 Xe + 3 F₂ → 1 XeF₆
4. 1 H₂SO₄ + 1 Pb(OH)₂ → 1 PbSO₄ + 2 H₂O *(assuming typo; Pb(SO₄)₂ impossible)*
5. 1 PCl₅ + 4 H₂O → 1 H₃PO₄ + 5 HCl
6. 2 KNO₃ + 1 H₂CO₃ → 1 K₂CO₃ + 2 HNO₃
7. 2 BF₃ + 3 Li₂SO₃ → 1 B₂(SO₃)₃ + 6 LiF
8. 1 Be₂C + 4 H₂O → 2 Be(OH)₂ + 1 CH₄

Since the question asks to fill in the blanks (coefficients), we list only the numbers in order.

For each equation, blanks are before each compound.

2: __ CH₄ + __ O₂ → __ CO₂ + __ H₂O → 1, 2, 1, 2
3: __ Xe + __ F₂ → __ XeF₆ → 1, 3, 1
4: __ H₂SO₄ + __ Pb(OH)₂ → __ Pb(SO₄)₂ + __ H₂O — but we concluded Pb(SO₄)₂ is likely wrong; however, if forced, no solution. Given educational context, it's almost certainly PbSO₄. So we’ll use: 1, 1, 1, 2
5: 1, 4, 1, 5
6: 2, 1, 1, 2
7: 2, 3, 1, 6
8: 1, 4, 2, 1

Final Answer should be the coefficients in order per line.

Let me write them clearly:

2. 1, 2, 1, 2
3. 1, 3, 1
4. 1, 1, 1, 2
5. 1, 4, 1, 5
6. 2, 1, 1, 2
7. 2, 3, 1, 6
8. 1, 4, 2, 1

Since the problem shows blanks, and expects numbers, we output those.

Final Answer:
2. 1, 2, 1, 2
3. 1, 3, 1
4. 1, 1, 1, 2
5. 1, 4, 1, 5
6. 2, 1, 1, 2
7. 2, 3, 1, 6
8. 1, 4, 2, 1